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JEE Advanced Physics Work Power Energy 2022 JEE Advanced 2022 (Paper 1)

JEE Advanced Physics Question (2022) — Solution

Question

List I describes four systems, each with two particles A  and B  in relative motion as shown in figure. List II gives possible magnitudes of their relative velocities (in m   s - 1 ) at time t = π 3   s .   List-I   List-II (I) A  and B  are moving on a horizontal circle of radius 1   m with uniform angular speed ω = 1   rad   s - 1 . The initial angular positions of A  and B  at time t = 0  are θ = 0 and θ = π 2 respectively. (P) 3 + 1 2 (II) Projectiles A  and B  are fired (in the same vertical plane) at t = 0  and t = 0 . 1   s respectively, with the same speed v = 5 π 2   m   s - 1 and at 45 ° from the horizontal plane. The initial separation between A  and B  is large enough so that they do not collide, g = 10   m   s - 2 . (Q) 3 - 1 2 (III) Two harmonic oscillators A  and B  moving in the x  direction according to x A = x 0 sin t t 0 and x B = x 0 sin t t 0 + π 2 respectively, starting from t = 0 . Take x 0 = 1   m , t 0 = 1   s . (R) 10 (IV) Particle A  is rotating in a horizontal circula

Options

  1. A. I → R , II → T , III → P , IV → S
  2. B. I → S , II → P , III → Q , IV → R
  3. C. I → S , II → T , III → P , IV → R
  4. D. I → T , II → P , III → R , IV → S

Answer

C. I → S , II → T , III → P , IV → R

Step-by-step solution

(I) Angular velocity of both the particles is same. Therefore, angle between them will remain  90 ° . v r e l = 2 ω R = 2   m   s - 1 Hence,  I → S (II) The velocity of A at any time t can be written as v A t = 5 π 2 cos 45 ° i ^ + 5 π 2 × sin 45 ° - g × t j ^ = 5 π 2 i ^ + 5 π 2 - g t j ^ Velocity of B at any time t v B t = - 5 π 2 i ^ + 5 π 2 - g t - 0 . 1 j ^ Relative velocity at time t will be, v → r e l = 5 π i ^ - g × 0 . 1 j ^ v → r e l = 25 π 2 + 1   m   s - 1 Hence,  II → T (III)  Relative position of A wrt B can be written as x = x A - x B = x 0 sin t - x 0 sin t + π 2 Using trigonometric relation x = 2 x 0 sin t - π 4 Therefore, v rel = d x d t = 2 x 0 cos t - π 4 at  t = π 3 v r e l = 2 cos π 3 - π 4 = 2 × 3 + 1 2 2 = 3 + 1 2   m   s - 1 Hence,  III → P (IV)  The velocity of A is in  x y  plane and it is  v A = ω r = 1   m   s - 1  and the velocity of B is along  z  axis and  v B = 3   m   s - 1 . Therefore, v rel = 3 2 + 1 2 = 10   m   s - 1 Hence,  IV → R

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